Symmetry and degeneracy

Symmetry and degeneracy are related as a result of the non-relativistic Hamiltonian being invariant under symmetry operations.

Consider the stationary state of a quantum mechanical system that is described by a set of linearly independent wavefunctions , which satisfies the time-independent Schrodinger equation . Subjecting both sides of the equation to the symmetry operation , where  are elements of a group , we have . Using eq53 and noting that  is a scalar, we obtain

which implies that  is an eigenfunction of , i.e. .

We say that the set of symmetry operations form the group of the Hamiltonian. If  is -fold degenerate, any linear combination of the subset of wavefunctions associated with the degenerate eigenvalue will be a solution of eq54. We can then write:

According to the closure property of ,

So,

Since  is a linear operator,

Subtracting eq59 from eq57, we have . If the wavefunctions are linearly independent, all coefficients of  must be zero, i.e.

With reference to eq55 through eq57,  and range from to . Thus, eq60 implies that  and are entries of the square matrices  and respectively, i.e.

Comparing eq61 with eq58, the matrices  and  multiply the same way as the symmetry operations  and  and therefore form a representation of . Such a representation has a dimension equal to  and may be reducible or irreducible. However, the following analysis shows that it is irreducible, assuming that there is no accidental degeneracy, which happens when two eigenvalues are the same even though their corresponding eigenfunctions transform differently under the symmetry operations of .

 

Question

What is an example of accidental degeneracy?

Answer

The hydrogenic wavefunctions and , which describe the -orbital and -orbitals respectively, obviously have different symmetry and may transform according to different irreducible representations, e.g. the -orbital transforms according to the totally symmetric irreducible representation, while the -orbitals may not. Therefore, they may have different energies. However, and  are degenerate wavefunctions for a particular value of . In fact, the number of degenerate energy levels of the hydrogen atom is  instead of , which is expected from a non-relativistic Hamiltonian with a central potential. This implies that the symmetry operations of the group of the Hamiltonian is incomplete. i.e. there are additional symmetries of the system that have not been taken into account when forming the group of the Hamiltonian.

 

To show that  is irreducible, we express eq55 as

where .

If  is reducible, then it may be in block-diagonal form. For example, if ,

Substituting eq61b in eq61a, and  are linear combinations of the subset with elements  and , while  and are linear combinations of the subset with elements  and . If so, the eigenvalues corresponding to the different subsets could be different (e.g. when one subset is one-dimensional and the other is two-dimensional), which would contradict our original assumption that  is -fold degenerate.

If is not initially in block-diagonal form, it can be converted into a block-diagonal matrix  via a similarity transformation, where . We can then rewrite eq61a as

where  is the basis for the transformed representation.

Since is in block-diagonal form, we arrive at the same conclusion as before. Therefore, assuming that there is no accidental degeneracy, a representation that is generated by a set of orthogonal degenerate wavefunctions is irreducible and we called the set of wavefunctions , basis wavefunctions of .

Question

1) Why is the number of basis functions needed to generate  equal to the dimension of ?

2) Can any set of functions, other than eigenfunctions of the Hamiltonian, be a set of basis functions for a representation of ?

3) Show that if  is a basis of a representation of a group, then any linear combination of  is a basis of a representation that is equivalent to .

Answer

1) Each  in of eq55 creates a column of matrix entries (see Q&A below for an illustration). So, the  matrix  is generated by  number of . The same logic applies to matrices  and . This implies that the number of linearly independent basis functions of a representation corresponds to the dimension of the representation. This set of linearly independent basis functions can be made orthogonal to one another using the Gram-Schmidt process. Therefore, the number of orthogonal basis functions of a representation corresponds to the dimension of the representation.

2) By inspecting eq55 through eq61, any set of linearly independent functions , not necessary eigenfunctions of the Hamiltonian, is a set of basis functions for a representation of  if  is transformed by a symmetry operation into a linear combination of . The converse is obviously also true, i.e. if  is a set of basis functions for a representation of , then . In this case, there is no requirement that the functions are degenerate and hence the representation generated can be either reducible or irreducible.

3) Since matrix multiplication is distributive,


Let’s rewrite eq55 as . Substituting this expression in the above equation and rearranging, gives


where .
Using eq61d and repeating the steps from eq55 through eq61, is also basis of a representation  of the group.
To Show that is equivalent to , let , which can be written as the following matrix equation:


where .

Using eq55 with in place of , we have , whose matrix equation is


where .

Multiplying eq61e by on the right, we have , which when substituted in eq61f gives


The matrix equation of eq55 with in place of is


Comparing eq61g with eq61h,  is related to by a similarity transformation. This implies that is equivalent to .

 

If is non-degenerate, the only eigenfunctions satisfying eq54 are , where is a constant. Therefore, . Normalising , we get , which means that . Consequently, the wavefunctions associated with non-degenerate eigenvalues transform according to one-dimensional irreducible representations with characters of +1 or -1.

 

Question

If , show how eq55 is related to the matrix representation .

Answer

The two degenerate wavefunctions are  and . We have

 

We can conclude from the above analysis that irreducible representations with dimension of one are called non-degenerate representations, while those with dimensions of more than one are known as degenerate representations.

 

Question

With regard to the  symmetry operation of the  point group, show that  and  transform together according to the two-dimensional irreducible representation .

Answer

With respect to the green line in the below diagram depicting the spherical coordinate system,  and are invariant to the symmetry operations of the  point group. If  and are angles before and after the  symmetry operation respectively, . In terms of the angle , we have .

The  and  wavefunctions in spherical coordinates are  and .

In other words, the symmetry operation  transforms  into a linear combination of  and (c.f. eq55). Similarly,

Therefore,  and  transform together according to the two-dimensional irreducible representation , with the coefficients of the basis functions  and  in eq62 and eq63 being the entries of the matrix .

 

Repeating the same procedure shown in the Q&A above for the rest of the symmetry operations of the  point group, we have

The transformation of the function  is determined by summing the separate transformations of the -orbitals  and .

 

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